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Question

A ball is projected horizontally from top of a $$80$$ m deep well with velocity $$10$$ m/s. Then particle will fall on the bottom at a distance of (all the collisions with the wall are elastic)

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A
5 m from A
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B
5 m from B
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C
2 m from A
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D
2 m from B
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Solution

The correct options are
B $$5$$ m from B
C $$2$$ m from A
Total time taken by the ball to reach at bottom $$=\sqrt{\dfrac{2H}{g}}=\sqrt{\dfrac{2\times 80}{10}}=4\ sec$$
Let time taken in one collision is $$t_{collision}$$. Then $$t_{collision} \times 10 = 7$$ since horizontal velocity is 10 m/s
$$ \Rightarrow t_{collision}=0.7$$ sec.
No. of collisions $$=\dfrac{4}{.7}=5\dfrac{5}{7}$$ (5th collisions from wall B)
Horizontal distance traveled in between $$2$$ successive collisions $$=7\:m$$
$$\therefore$$Horizontal distance traveled in $$5/7$$ part of collisions $$=$$ The 1st collision will be with wall B, 2nd collision with wall A, 3rd collision with wall B, 4th collision with wall A and  5th collision will be with wall B.  After 5th collision, the ball   will cover a distance of $$\dfrac{5}{7}\times 7=5 m$$ from wall B i.e it will have collision at a distance of 2m from wall A.

Physics

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